I'm a halfer, but think you did your math wrong when calculating the thirder view.

The thirder view is that the probability of an event happening is the experimenter's expectation of the proportion of awakenings where the event happened.

So for your setup, with k=2:

There are three possible outcomes: H, HT, and TT.

H happens in 50% of experiments, HT happens in 25% and TT happens in 25%.

When H happens there is 1 awakening, when HT happens there are 2 awakenings, and when TT happens there are 4 awakenings.

We'll imagine that the experiment is run 4 times, and that H happened in 2 of them, HT happened once, and TT happened once. This results in 2*1=2 H awakenings, 1*2=2 HT awakenings, and 1*4=4 TT awakenings.

Therefore, H happens in 2/(2+2+4)=25% of awakenings, HT happens in 25% of awakenings, and TT happens in 50% of awakenings.

The thirder view is thus that upon awakening Beauty's credence that the coin came up heads should be 25%.

What is you [sic] credence that in this experiment the coin was tossed k times and the outcome of the k-th toss is Tails?

Answering your question, the thirder view is that there was a 6/8=75% chance the coin was tossed twice, and a 4/6 chance that the second toss was a tails conditional on it being the case that two tosses were made.

Unconditionally, the thirder's credence is 4/8=50% chance that it is both true that the coin was tossed two times and that the second toss was a tails.

I've been trying to make this comment a bunch of times, no quotation from the post in case that's the issue:

No, a thirder would not treat those possibilities as equiprobable. A thirder would instead treat the coin toss outcome probabilities as a prior, and weight the possibilities accordingly. Thus H1 would be weighted twice as much as any of the individual TH or TT possibilities.

As simon has already said [LW(p) · GW(p)], your math is wrong because the cases aren't equiprobable. For k=2 you can fix this by doubling the cases of H since they're twice as likely as the others (so the proper state space is $\Omega =\{H,H,HT,HT,TT,TT,TT,TT\}$ with $p\left(TT\right)=\frac{1}{2}$.) For $k=3$ you'd have to quadruple H and double HT, which would give 4x H and 4x HT and 4x HTT and 8x TTT I believe, leading to $\frac{2}{5}$ probability of TTT. (Up from 1x H, 2x HT, 4x HTT, 8x TTT.) In general, I believe the probability of only Ts approaches 0, as does the probability of H.

Regardless, are these the right betting odds? Yup! If we repeat this experiment for any $k$ and you are making a bet every time you wake up, then these are the odds according to which you should take or reject bets to maximize profit. You can verify this by writing a simulation, if you want.

If you make the experiment non-repeating, then I think this is just a version of the presumptuous philospoher argument which (imo) shows that you have to treat logical uncertainty differently from randomness (I addressed this case here [LW(p) · GW(p)]).